Probability?

Question

In a tire factory, the quality control inspector examines a random sample of 15
tires. When more than one defective tire is found, production is halted. Suppose
the defective rate is 6%. What is the probability that the production is halted?

Answer 1

The probability that production is halted is 1 - P(all items pass).

Since each item is likely to be good 94% of the time, the chance that everything passes is 0.94 x 0.94 x 0.94 x... x 0.94 (15 times). In other words, the chance that everything passes is 0.94^15.

If you calculate this out it comes to 0.395291799 or approximately 39.5% success rate.

That means the chance that at least one tire is defective and production must be halted is:
1 - 0.94^15

This is 0.604708201 or approximately 60.5% probability that the production line will have to be stopped. As others have noted, a failure rate like this would be unacceptable for a real-world company.

Edit: I just reread the question and realized that that production is stopped if there is *more* than 1 defect tire. My calculations were for *at least* 1 (in other words more than 0). To correct this, you would want to figure out the probability of exactly 1 failure and subtract this as well from the answer.

The chance that a selected item is bad and the other 14 are good is 0.06 * 0.94^14. But then you have 15 ways to have one item bad. This is 0.378470871, or about 37.85%. Subtract that from the prior answer I gave.

In summary:
P(none are bad) = 0.94^15 = 0.3953
P(exactly one is bad) = 15 * 0.06 * 0.94^14 = 0.3785

1 - P(all are good) - P(exactly one is bad, 14 are good)
1 - 0.3953 - 0.3785
= 0.2262
= 22.62%

Answer 2

i've got faith my argument could no longer be properly won. I say the possibility is 50% permit a ??, permit b ?? and randomly opt for the values for a and b. As already suggested, for a ? 0, P( a < b²) = a million, that's trivial. in basic terms rather much less trivial is the theory P(a < 0 ) = a million/2 and consequently P( a < b² | a ? 0) = a million and P( a < b² ) ? a million/2 Now evaluate what happens whilst a > 0 For a > 0, whilst that's undemanding to tutor there's a non 0 probability for a finite b, the shrink, the possibility is 0. a < b² is reminiscent of asserting 0 < a < b², undergo in strategies we are in basic terms finding at a > 0. If this a finite era on an countless line. The probability that a is an factor of this era is 0. P( a < b² | a > 0) = 0 As such we've a finished probability P( a < b² ) = P( a < b² | a ? 0) * P(a ? 0) + P( a < b² | a > 0) * P(a > 0) = a million * a million/2 + 0 * a million/2 = a million/2 undergo in strategies, that's because of the countless instruments. no count what form of era you draw on paper or on a working laptop or pc you will discover a finite probability that seems to physique of strategies a million. yet that's because of the finite random selection turbines on the pc and if we had this question asked with finite values there may well be a a answer greater desirable than 50%. i do no longer advise to be condescending, yet please clarify why employing the Gaussian to approximate a uniform distribution is a robust theory? are not countless numbers exciting. Cantor whilst mad working with them! :)

Answer 3

This is a binomial probability distribution because there is a certain number of trials (n = 15) and at each trial the probability of "success" (not defective, p = 0.94) and of "failure" (q = 0.06) is the same. So the formula for probability of a particular number of successes (r) is
(nCr) x p^r x q^(n-r)

There have to be 14 or 15 successes, otherwise the line is halted. So use the above formula to work out the probabilities of 14 or 15 successes, add them together and you have the probability that the line is NOT halted. Subtract this from 1 and you have the answer. I make it about 22.6%, hope you do the same!

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Answer 4

This is the binomial distribution with p = 0.06, and q = 0.94, N = 15, and n = 0 and 1.

Probability of zero defects chosen:
P(0) = C(15, 0)(0.06)^0(0.94)^15 = 0.395

Probability of exactly one defect chosen:
P(1) = C(15, 1)(0.06)^1(0.94)^14 = 0.378

Probability of two or more defects = 1 - P(0) - P(1)
= 1 - 0.395 - 0.378
= 0.227

If the defect rate is really 6% (unbelievably high for any manufacturing company), the probability of halting production is 22.7%.

Answer 5

The defective production rate is 6%, that is 1 in 16.6666. So divide 15 tires by the probability rate so if one defective tire stopped production then the probability would be15 divided by 16.6666 = 90% but as 2 or more must be defective to stop production the probablility is half that, ie 45%.

Answer 6

This is a binomial probability question with the chance of success being 0.06, n the sample size 15,

We want to know the p(X>=2)
That is the same as 1 - P(x=0) - P(x=1)
1 - 0.3953 - 0.3785 = 0.2262 round to 0.226

Answer 7

any company with such a shoddy quality control process will get sued out of business. so 100% probability that production will be halted. Permanently.

Answer 8

.9

Answer 9

75%